Using @[array, of, items] vs [NSArray arrayWithObjects:]

Question

Is there a difference between

NSArray *myArray = @[objectOne, objectTwo, objectThree];

and

NSArray *myArray = [NSArray arrayWithObjects:objectOne, objectTwo, objectThree, nil];

Is one preferred over the other?

Nervine

possible duplicate of Difference between @[] and [NSArray arrayWithObjects:] – Possessory

Sorry, I even searched google and here. Maybe the symbols were messing up the search that is why i wasn't getting anything relevant. – Nervine

It's not meant to be a recrimination. Actually, that comment is created by the system on my behalf when I vote to close. – Possessory

Answer 1

They are almost identical, but not completely. The Clang documentation on Objective-C Literals states:

Array literal expressions expand to calls to +[NSArray arrayWithObjects:count:], which validates that all objects are non-nil. The variadic form, +[NSArray arrayWithObjects:] uses nil as an argument list terminator, which can lead to malformed array objects.

So

NSArray *myArray = @[objectOne, objectTwo, objectThree];

would throw a runtime exception if objectTwo == nil, but

NSArray *myArray = [NSArray arrayWithObjects:objectOne, objectTwo, objectThree, nil];

would create an array with one element in that case.

Fourhanded

+1 This is the correct answer; the difference is subtle, but nevertheless the two are different. – Tolle

I'm not sure what "malformed array objects" means, but the arrays would be perfectly fine, just not the length you thought they would be. Also, I sometimes intentionally exploit the nil termination behavior, to have an optional last element that is only included if it is not nil. – Eliseo

Answer 2

No. At compile time the @[...] literals will be changed to arrayWithObjects:

The only difference is that @[...] is only supported in newer versions of the LLVM compiler.

Chastise

No it is arrayWithObjects:count: – Eliseo